3.1199 \(\int \sqrt [4]{a-b x^4} \, dx\)

Optimal. Leaf size=83 \[ \frac {1}{2} x \sqrt [4]{a-b x^4}-\frac {\sqrt {a} \sqrt {b} x^3 \left (1-\frac {a}{b x^4}\right )^{3/4} F\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 \left (a-b x^4\right )^{3/4}} \]

[Out]

1/2*x*(-b*x^4+a)^(1/4)-1/2*(1-a/b/x^4)^(3/4)*x^3*(cos(1/2*arccsc(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccsc
(x^2*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arccsc(x^2*b^(1/2)/a^(1/2))),2^(1/2))*a^(1/2)*b^(1/2)/(-b*x^4+a)^(3/4
)

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Rubi [A]  time = 0.03, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {195, 237, 335, 275, 232} \[ \frac {1}{2} x \sqrt [4]{a-b x^4}-\frac {\sqrt {a} \sqrt {b} x^3 \left (1-\frac {a}{b x^4}\right )^{3/4} F\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 \left (a-b x^4\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^4)^(1/4),x]

[Out]

(x*(a - b*x^4)^(1/4))/2 - (Sqrt[a]*Sqrt[b]*(1 - a/(b*x^4))^(3/4)*x^3*EllipticF[ArcCsc[(Sqrt[b]*x^2)/Sqrt[a]]/2
, 2])/(2*(a - b*x^4)^(3/4))

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 232

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(3/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rule 237

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[(x^3*(1 + a/(b*x^4))^(3/4))/(a + b*x^4)^(3/4), Int[1/(x^3*
(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ[{a, b}, x]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \sqrt [4]{a-b x^4} \, dx &=\frac {1}{2} x \sqrt [4]{a-b x^4}+\frac {1}{2} a \int \frac {1}{\left (a-b x^4\right )^{3/4}} \, dx\\ &=\frac {1}{2} x \sqrt [4]{a-b x^4}+\frac {\left (a \left (1-\frac {a}{b x^4}\right )^{3/4} x^3\right ) \int \frac {1}{\left (1-\frac {a}{b x^4}\right )^{3/4} x^3} \, dx}{2 \left (a-b x^4\right )^{3/4}}\\ &=\frac {1}{2} x \sqrt [4]{a-b x^4}-\frac {\left (a \left (1-\frac {a}{b x^4}\right )^{3/4} x^3\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1-\frac {a x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{x}\right )}{2 \left (a-b x^4\right )^{3/4}}\\ &=\frac {1}{2} x \sqrt [4]{a-b x^4}-\frac {\left (a \left (1-\frac {a}{b x^4}\right )^{3/4} x^3\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-\frac {a x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{x^2}\right )}{4 \left (a-b x^4\right )^{3/4}}\\ &=\frac {1}{2} x \sqrt [4]{a-b x^4}-\frac {\sqrt {a} \sqrt {b} \left (1-\frac {a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 \left (a-b x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 47, normalized size = 0.57 \[ \frac {x \sqrt [4]{a-b x^4} \, _2F_1\left (-\frac {1}{4},\frac {1}{4};\frac {5}{4};\frac {b x^4}{a}\right )}{\sqrt [4]{1-\frac {b x^4}{a}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^4)^(1/4),x]

[Out]

(x*(a - b*x^4)^(1/4)*Hypergeometric2F1[-1/4, 1/4, 5/4, (b*x^4)/a])/(1 - (b*x^4)/a)^(1/4)

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-b x^{4} + a\right )}^{\frac {1}{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

integral((-b*x^4 + a)^(1/4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-b x^{4} + a\right )}^{\frac {1}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

integrate((-b*x^4 + a)^(1/4), x)

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \left (-b \,x^{4}+a \right )^{\frac {1}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b*x^4+a)^(1/4),x)

[Out]

int((-b*x^4+a)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-b x^{4} + a\right )}^{\frac {1}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

integrate((-b*x^4 + a)^(1/4), x)

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mupad [B]  time = 1.07, size = 38, normalized size = 0.46 \[ \frac {x\,{\left (a-b\,x^4\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ \frac {b\,x^4}{a}\right )}{{\left (1-\frac {b\,x^4}{a}\right )}^{1/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x^4)^(1/4),x)

[Out]

(x*(a - b*x^4)^(1/4)*hypergeom([-1/4, 1/4], 5/4, (b*x^4)/a))/(1 - (b*x^4)/a)^(1/4)

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sympy [C]  time = 0.97, size = 39, normalized size = 0.47 \[ \frac {\sqrt [4]{a} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x**4+a)**(1/4),x)

[Out]

a**(1/4)*x*gamma(1/4)*hyper((-1/4, 1/4), (5/4,), b*x**4*exp_polar(2*I*pi)/a)/(4*gamma(5/4))

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